Complete the square to solve for $x$. $x^{2}-8x+12 = 0$
Answer: Begin by moving the constant term to the right side of the equation. $x^2 - 8x = -12$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-8$ , half of it would be $-4$ , and squaring it gives us ${16}$ $x^2 - 8x { + 16} = -12 { + 16}$ We can now rewrite the left side of the equation as a squared term. $( x - 4 )^2 = 4$ Take the square root of both sides. $x - 4 = \pm2$ Isolate $x$ to find the solution(s). $x = 4\pm2$ So the solutions are: $x = 6 \text{ or } x = 2$ We already found the completed square: $( x - 4 )^2 = 4$